Numbers between $1 - 1000$ which leave no remainder when divided by $4$ and divided by $6$ but not by $21$?

I tried $$\frac{1000}{12} = 83 - \frac{83}{21} = 83-3 = 80$$

Am I correct? Can someone please explain to me how it works?

  • $\begingroup$ You should explain your reasoning when you post an attempt since it helps users identify any errors you may have made. This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Feb 26 at 10:19

If a number is divisible by both $4$ and $6$, then it is divisible by $\operatorname{lcm}(4, 6) = 12$. The number of multiples of $12$ that are at most $1000$ is $$\left\lfloor \frac{1000}{12} \right\rfloor = 83$$ where $\lfloor x \rfloor$ is the greatest integer less than $x$.

From these, we must subtract those numbers that are also divisible by $21$. Those numbers are divisible by $\operatorname{lcm}(4, 6, 21) = \operatorname{lcm}(12, 21) = 84$. The number of multiples of $84$ that are at most $1000$ is $$\left\lfloor \frac{1000}{84} \right\rfloor = 11$$ Hence, the number of positive integers less than or equal to $1000$ that are divisible by both $4$ and $6$ but not divisible by $21$ is $$\left\lfloor \frac{1000}{12} \right\rfloor - \left\lfloor \frac{1000}{84} \right\rfloor = 83 - 11 = 72$$


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